Choose the correct match-
A)Gama line in Lyman series in H--UV
B)Beta line in Balmer series in He⁺--UV
C)Delta line in Balmer series in H--visible
D)Delta line in Paschen series in H--infrared
Answer is all the options are correct. I have already asked this question and expert told me how to solve it but his question is new to me and I have no idea about what the wavelength must be to be in which series. So pls solve them. thanks!

Lyman series is transition from n=1 to higher orbits.
Alpha line is the first transition, i.e from n=2 to n=1, beta is from n=3 to n=1. gamma is from n=4 to n=1 and so on.....
Balmer ​series is transition from n=2 to higher orbits.
​Alpha line is the first transition, i.e from n=3 to n=2, beta is from n=4 to n=2, gamma is from n=5 to n=2 and so on......
Paschen series is transition from n=3 to higher orbits.
​​Alpha line is the first transition, i.e from n=4 to n=3, beta is from n=5 to n=3, gamma is from n=6 to n=3 and so on......

We can find the wavelength corresponding to these transitions by applying the formula:
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\mathrm{R}(\frac{1}{\mathrm{n}{\text{'}}^2}-\frac{1}{{\mathrm{n}}^2}) \\ \mathrm{n}\text{'}=\mathrm{lower} \mathrm{enery} \mathrm{shell} \\ \mathrm{n}= \mathrm{higher} \mathrm{energy} \mathrm{shell} \end{gathered}$$

a) Lyman-alpha is wavelength corresponding to  n=2 to n=1
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}(\frac{1}{1^2}-\frac{1}{2^2}) \\ =1.09\times {10}^7 {\mathrm{m}}^{-1} \times (0.75)=8.175\times {10}^6 {\mathrm{m}}^{-1} \\ \mathrm{\lambda }=0.122\times {10}^{-6} \times {10}^9 \mathrm{nm}=0.122\times {10}^3 = 122 \mathrm{nm} \end{gathered}$$
See that this Lyman series lines is in UV region (less tan 400 nm).

b) Beta line in Balmer series in He⁺ is line corresponding to n=4 to n=2.
Rydberg constant, R for He is 4 times (Z²​ and Z=2 for He) the Rydberg constant of Hydrogen. 
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{He}}(\frac{1}{2^2}-\frac{1}{4^2}) \\ =\mathbf{4}\times 1.09\times {10}^7 {\mathrm{m}}^{-1} \times (0.1875)=8.175\times {10}^6 {\mathrm{m}}^{-1} \\ \mathrm{\lambda }=0.122\times {10}^{-6} \times {10}^9 \mathrm{nm}=0.122\times {10}^3 = 122 \mathrm{nm} \\ \mathrm{If} \mathrm{for} \mathrm{hydrogen} \mathrm{then} \mathrm{wavelength} \mathrm{would} \mathrm{have} \mathrm{been} 4\times 122 = 488 \mathrm{nm} \mathrm{that} \mathrm{falls} \mathrm{in} \mathrm{visible} \mathrm{region}. \end{gathered}$$
Beta line in Balmer series in H is in visible region but for He⁺ the value of Z² is 4, so the wavelength is four times shorter than the Balmer series of H atom and hence it lies in UV region.

c) Delta line in Balmer series in H can be calculated similarly:
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}(\frac{1}{2^2}-\frac{1}{6^2}) \\ \mathrm{\lambda } =410.2 \mathrm{nm} (\mathrm{visible} \mathrm{region}) \end{gathered}$$

d) Delta line in Paschen series in H
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}={\mathrm{R}}_{\mathrm{H}}(\frac{1}{2^2}-\frac{1}{7^2}) \\ \mathrm{\lambda } =1005 \mathrm{nm} (\mathrm{infra} \mathrm{red} \mathrm{region}) \end{gathered}$$

Hope that's clear now.